---
title: "150 km line with mid-point shunt capacitor"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:150km-mid-shunt-capacitor"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/150km-mid-shunt-capacitor
site: BCT Engineering
---

# 150 km line with mid-point shunt capacitor

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2070 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/150km-mid-shunt-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 150 km long, 50 Hz, 132 kV, 3-phase overhead line with series impedance of 0.1 + j0.3 Ω/m and shunt admittance of 3 × 10⁻⁶∠90° S/km is feeding a load of 35 MW at 0.9 p.f. lagging. Determine the transmission line efficiency when a shunt capacitor of susceptance 4 × 10⁻⁶∠90° S is connected at the middle of the line.

## Short answer

With a shunt capacitor of susceptance 4×10⁻⁶ S at the midpoint, the transmission efficiency is approximately 96.7%.

## Answer

**Assumptions:** the series impedance is taken as $0.1 + j0.3\ \Omega$/**km** (Ω/m is a misprint). The load is at 132 kV. The line is modelled as two 75 km nominal-π sections with the capacitor at the junction.

#### Section constants (75 km each)
$$
\begin{aligned}
Z_h &= (0.1 + j0.3)\times75 = 7.5 + j22.5\ \Omega \\
Y_h &= j3\times10^{-6}\times75 = j2.25\times10^{-4}\ \text{S},\quad Y_h/2 = j1.125\times10^{-4}\ \text{S}
\end{aligned}
$$
At the middle bus, the two half-section admittances and the capacitor add: $Y_{mid} = j(2.25\times10^{-4} + 4\times10^{-6}) = j2.29\times10^{-4}$ S.

```text
Vs    Zh          Vm          Zh      Vr
o---/\/\/---+-----+-----+---/\/\/---o
|           |     |     |           |
Yh/2      Yh/2   Ysh   Yh/2       Yh/2
|           |     |     |           |
o-----------+-----+-----+-----------o
```

#### Receiving-end section
$$
\begin{aligned}
V_R &= 132/\sqrt3 = 76\,210\angle0^\circ\ \text{V} \\
I_R &= \frac{35\times10^6}{\sqrt3\times132\,000\times0.9} = 170.09\angle-25.84^\circ\ \text{A} \\
I_1 &= I_R + j\frac{Y_h}{2}V_R = 170.09\angle-25.84^\circ + j8.57 = 166.54\angle-23.19^\circ\ \text{A} \\
V_m &= V_R + I_1Z_h = 78\,889\angle2.15^\circ\ \text{V}
\end{aligned}
$$

#### Middle bus and sending section
$$
\begin{aligned}
I_{mid} &= Y_{mid}V_m = 18.07\angle92.15^\circ\ \text{A} \\
I_2 &= I_1 + I_{mid} = 159.64\angle-17.32^\circ\ \text{A} \\
V_S &= V_m + I_2Z_h = 81\,270\angle4.25^\circ\ \text{V} \\
I_S &= I_2 + j\frac{Y_h}{2}V_S = 156.51\angle-14.20^\circ\ \text{A}
\end{aligned}
$$
Sending-end line voltage $= 140.76$ kV.

#### Efficiency
$$
\begin{aligned}
S_S &= 3V_SI_S^* = 36.20 + j12.08\ \text{MVA} \\
\eta &= \frac{35}{36.20}\times100 = 96.69\%
\end{aligned}
$$
(Check: $3(|I_1|^2 + |I_2|^2)(7.5) = 1.20$ MW loss.)

A $4\times10^{-6}$ S capacitor supplies only about $V^2B \approx 0.07$ MVAr, so it hardly changes the result (without it, $\eta$ = 96.68 %).

**Answer: Transmission efficiency ≈ 96.7 %.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the 150 km line modelled in this solution?
   A. One nominal-pi section of 150 km
   B. Two 75 km nominal-pi sections with the capacitor at the junction
   C. Two T sections with the capacitor at the sending end
   D. A short line ignoring admittance
   Answer: B. The line is split into two 75 km nominal-pi sections and the capacitor sits at the middle bus. (Key point: Model used)

2. What shunt admittance appears at the middle bus?
   A. Only the capacitor admittance 4e-6 S
   B. The two half-section admittances Y_h/2 each only
   C. Y_h/2 only plus the capacitor
   D. The two Y_h/2 halves (total Y_h) plus the capacitor susceptance
   Answer: D. The end halves of both adjacent pi sections join at the middle bus, giving Y_h = 2.25e-4 S, plus 4e-6 S, so 2.29e-4 S. (Key point: Admittance at the middle bus)

3. Starting from the load, what is the correct sequence of calculation?
   A. I_R, then I_1 = I_R + (Y_h/2)V_R, then V_m = V_R + I_1 Z_h, then add middle-bus current and repeat
   B. V_S first, then work back to V_R
   C. Assume V_m = V_R and skip the second section
   D. Compute efficiency from the capacitor MVAr alone
   Answer: A. Working from the receiving end, each section adds its shunt current and then its series drop. (Key point: Order of calculation)

4. Why does the 4e-6 S capacitor hardly change the efficiency (96.69% vs 96.68%)?
   A. It supplies only about 0.07 MVAr, tiny compared with the line charging and load
   B. It is connected in series so it has no effect
   C. It is cancelled by the line inductance exactly
   D. Efficiency is independent of reactive power in all cases
   Answer: A. V^2 B is about 0.07 MVAr, so current and losses barely change. (Key point: Effect of the small capacitor)

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